Sorry for the necromancy (and I know it's unlikely the OP will see this), but this can be answered. First, @The Photon's answer has good points, so my aim is not competition, but completeness.
The filter can be calculated, but there are a few things to know, first:
- The impedance of a typical electrodynamic speaker has a bump at its resonance frequency, and an upward slope towards the high frequencies. Below is a possible representation of a (rather simplified version of a) woofer:
The value of the impeance is set to be the one at the bottom of the "bell", or closer to DC (but not DC, that would fry the coil), or @1 kHz. For subwoofers I've also seen defined at 100 Hz, since their resonant frequency can be ~20~30 Hz. For this one, the values read 8.94 Ω @1 kHz, and 8.5 Ω @100 Hz -- both are close to a standard 8 Ω speaker.
- The output impedance of the amplifier needs to be low, not 50 Ω -- that would be a typical value of a signal generator. One of the reasons has to do with the point above: it will provide enough damping such that the variation of the magnitude of the signal will be fairly constant across the spectrum. For example, here's how the output of the amplifier varies with the same woofer, but with 0.1 Ω output impedance vs 1 Ω (I'll let you guess which one is which):
- Not lastly, passive filters suffer from one drawback: every component is influenced by its neighbour (and beyond), which makes them a one-shot deal. Calculating them means imposing a fixed value for the I/O impedances/resistances, and once calculated, their values remain fixed. Any change in the I/O results either in a recalculation of the whole filter, or an acceptance that the response will change.This is the only thing that needs mentioning in @ThePhoton's answer: the resistances will influence the frequency. To show what I mean, a 4th order LC filter with \$R_i\$ and \$R_o\$ as I/O resistances, will have this transfer function:
$$\begin{align}
H(s)&=\dfrac{\dfrac{1}{L_1L_2C_1C_2}}{s^4+\left(\dfrac{1}{R_oC_2}+\dfrac{R_i}{L_1}\right)s^3+\left(\dfrac{R_i}{R_oL_1C_2}+\dfrac{1}{L_2C_2}+\dfrac{1}{L_2C_1}+\dfrac{1}{L_1C_1}\right)s^2+\left[\dfrac{1}{R_oC_1C_2}\left(\dfrac{1}{L_1}+\dfrac{1}{L_2}\right)+\dfrac{R_i}{L_1L_2}\left(\dfrac{1}{C_1}+\dfrac{1}{C_2}\right)\right]s+\dfrac{R_i+R_o}{R_o}\dfrac{1}{L_1L_2C_1C_2}} \tag{1} \\
\omega_p&=\sqrt{\dfrac{R_i+R_o}{R_o}\dfrac{1}{L_1L_2C_1C_2}} \tag{2}
\end{align}$$
As you can see, (2) has both \$R_i\$ and \$R_o\$ in its formula, and they're even spread in the rest of \$H(s)\$'s denominator.
With these in mind, you can calculate the filter yourself. First, you need the prescription, from which two things are known: \$f_p=100\;\text{Hz}\$ (a typical value for woofers) and \$N=4\$. You also need the I/O impedances, and here comes the first minor hurdle: the audio amplifiers have, typically, a low impedance, and if it is 10x (or more) lesser than \$R_o\$, it can be considered a short (it can be omitted, set to zero). Sure enough, in (1), setting \$R_i=0\$ will cancel a few terms and will make \$\omega_p\$ equal to the numerator, which means there will be no attenuation. For this case, let's consider \$R_i=0.1\;\Omega\$, a not so great, but not so bad value, either. Then, the speaker can be considered 8 Ω.
Not lastly, the type of the filter. Since this is audio, the flatness of the passband is what matters most (a), but also the cross-over (b):
a) Out of all the designs for classical filters, the one with the maximum flatness in the passband is inverse Chebyshev (type II), flatter than Butterworth, but this is a pole-zero filter, which makes the design not only a bit more complicated (certainly not impossible), but the zeroes might affect the transient response in an unwanted way. Therefore an all-pole filter needs to be considered, and here ...
b) ... you have two choices: Butterworth (maximum flatness for all-pole filters), or Linkwitz-Reily. If you have other speakers connected to this, and they're using the woofer's \$f_p\$ to continue, upwards, in frequency, then you need L-R, since these provide a -6 dB cutoff which, when interconnected, results in a flat spectrum; B. filters will have a bump in magnitude, which means you'll hear amplified sounds at the cross-over. To show what I mean, here are two B. filters, V(B*), one lowpass and one highpass, and how their sum looks like, vs the same thing but L-R:
So, I'll choose a L-R. To solve, you'll need the transfer function, and L-R filters are nothing but B in disguise: a 4th order L-R is actually two, identical B filters, cascaded. Therefore, start with the lowpass prototype transfer function:
$$B_2(s)=\dfrac{1}{s^2+\sqrt2 s+1}\quad\rightarrow\quad B_2(s)^2=LR_4(s)=\dfrac{1}{s^2+2\sqrt2 s^3+4s^2+2\sqrt2 s+1} \tag{3}$$
This must be multiplied by the voltage divider ratio, \$R_o/(R_i+R_o)\$, to give the numeric transfer function. We already have (1), we found out (3), so now equate the coefficients of \$s\$ for both transfer function denominators to obtain a system of 4 polynomials, then solve for \$L_1,\;L_2,\;C_1,\;C_2\$. Depending on what solver you're using you may get more than one solution. I used wxMaxima and I got (with fpprintprec:4):
[l2=7.556,l1=15.16,c2=0.0443,c1=0.1995]
[l2=0.1596,l1=0.03544,c2=18.95,c1=9.445]
[l2=12.02,l1=0.06697,c2=0.09361,c1=13.44]
[l2=11.38-0.6275*%i,l1=0.07049-0.003935*%i,c2=0.004918*%i+0.08811,c1=0.7844*%i+14.23]
[l2=0.6275*%i+11.38,l1=0.003935*%i+0.07049,c2=0.08811-0.004918*%i,c1=14.23-0.7844*%i]
[l2=10.75,l1=0.07489,c2=0.08372,c1=15.02]
[l2=11.42-11.3*%i,l1=0.03535*%i+0.03552,c2=0.0444-0.04419*%i,c1=14.12*%i+14.27]
[l2=11.3*%i+11.42,l1=0.03552-0.03535*%i,c2=0.04419*%i+0.0444,c1=14.27-14.12*%i]
4 out of 8 solutions are complex, so they are not needed, which leaves 4 valid solutions. The advantage is that you can cherry-pick any solution that is more favourable. In this case, the 2nd solutions involves low values for inductors, which is desirable since those shouldn't be wound on a core to avoid nonlinearities, which means they will turn out to be bulky.
All that's left now is to scale the elements and test the result:
The attenuation @\$f_p\$ is less than 6 dB because the input is not zero, but finite. Also, the green numbers represent the values for the elements. The inductors are ridiculously small, but the capacitors are now too large (tens of mF), so you may want to test out the other values.
And now, the final moment, when you add the "real" load:
The black traces are all the four solutions with 0.1 Ω input resistance, the blue ones are with 1 Ω input resistance, to show what happens if the amplifier a too large output impedance, and the red ones are the reference (with a pure 8 Ω output resistance, same 0.1 Ω input).
In all the cases, the blue traces are distorted due to the low damping provided by the source (the 1 Ω amplifier output). Out of all the black traces, the only one that almost overlaps with the reference red trace is the 3rd from the top, and that's the one that has the lowest values for the inductances (fractions of mH), but huge capacitors (tens of mF!). The bottom plot has mH L and μF C, while the other two have them mixed. You could do it by using series polarized caps, connected - to -, for example, but their capacitance will halven, and you'll need quite a few of these in parallel to make the 15 mF and 30 mF. Or you could accept one of the other responses with a "what-can-you-do" shrug. Or you could make your own amplifier the incorporates the output filter in its feedback and, provided you take care of the phase, you should be end up with an output like the reference. Choices.
